Question

An electric current is passed through two electrolytic cells connected in series, one containing $Ag{NO}_3$(aq) and other $H_2{SO}_4$(aq). What volume of $O_2$ measured at ${25}^{o}C$ and 750 mm Hg would be liberated in litre from $H_2{SO}_4$, if $8\times {10}^{22}$ ions of $A{g}^{+}$ are deposited from $Ag{NO}_3$ solution? (Round off to integer)

Answer 1

Anode:

$2H_{2}O\to 4H^{+}+O_2+4e^{-}$
Cathode:
$4A{g}^{+}+4e^{-}\to 4Ag$
From 4mole of Ag we get 1 mole of $O_2$
1 mole of Ag $\Rightarrow$ 1/4 mole of $O_2$

Eq. of $O_2$ = Eq of $Ag$
$n_{Ag}=\frac{8\times {10}^{22}}{6.023\times {10}^{23}}=0.133$
$n_{O_2}=\frac{0.133}{4}=0.033$

Again using $PV=nRT$; $V_{O_2}=\frac{0.033\times 0.0821\times 298}{\left(\frac{750}{760}\right)atm}=0.823Litre$
Round off integer value of the volume of Oxygen is 1.

Hence, the correct option is 1.